poj 3176 Cow Bowling(区间dp)

题目链接:http://poj.org/problem?id=3176

思路分析：基本的DP题目；将每个节点视为一个状态，记为B[i][j], 状态转移方程为 B[i][j] = A[i][j] + Max( B[i+1][j], B[i+1][j+1] );

代码如下： 

#include <stdio.h>
 
const int MAX_N = 350 + 10;
int A[MAX_N][MAX_N], B[MAX_N][MAX_N];

int Max( int a, int b ) { return a > b ? a : b; }
int dp( int i, int j )
{
    if ( B[i][j] >= 0 )
        return B[i][j];

    return B[i][j] = A[i][j] + Max( dp(i+1, j), dp(i+1, j+1) );
}

int main()
{
    int n, ans;

    scanf( "%d", &n );
    for ( int i = 1; i <= n; ++i )
        for ( int j = 1; j <= i; ++j )
        {
            scanf( "%d", &A[i][j] );
            B[i][j] = -1;
        }

    for ( int k = 1; k <= n; ++k )
        B[n][k] = A[n][k];

    ans = dp( 1, 1 );
    printf( "%d
", ans );

    return 0;
}